Explanation for this chemistry question

[Pdentalstudent]

I am unsure of how to answer this question and I don't understand the answer

1. When the volume of a gas is decreased at constant temperature, the pressure increases because the molecules:
a) Move faster
b) Move slower
c) Become heavier
d) Become lighter
e) Strike a unit area of the container more often

The answer is e)
Doesn't decreasing the volume = decrease in the gas molecules. Wouldn't that lead to more dispersion of the gas molecules?

2. The following is a spontaneous oxidation-reduction reaction:
(Cr2O7^2-) + (14H^+) + (6I^-) + --> (2Cr^3+) + 7H2O + 3I2
Which of the following is the best reducing agent?
The answer is I- but I don't understand why H+ the best reducing agent


Thanks in advance

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[kevinrho]

Other SDN'ers will contribute and will probably have a better answer, but hopefully I will be able to give you an answer without confusing you.

Do you have chad's video? He explains it pretty good in Day 3 under gases. If I remember correctly he used the analogy of, 10 people (gas molecules) in a class room there isn't going to be a whole lot of pressure. However, if you move all 10 people into a small bathroom there will be more pressure because of the small volume. The gas molecules are now running into the wall more often.

Question 1.
Boyle's Law: Pressure is inversely proportional to Volume. When you decrease volume you will increase pressure.

According to your understanding, you think that decreasing volume will decrease the gas molecules. However, you are not letting any of the gas molecules out. As the example earlier (still 10 gas molecules, but in a smaller container).

hopefully i didn't confuse you! I know you weren't understanding why the reason was, I don't know if I was able to explain it, but I tried...

Question 2.

The answer is indeed I-. I- is being oxidized therefore it will be the reducing agent. I- = -1 charge however it was oxidized (loss of electron) and in the product it did not have a -1 charge anymore.

H+ is not being reduced or oxidized. It cannot be the oxidizing agent. H+ had charge of +1, and in the product it still had a charge of +1. O = -2, you need 2 H+ in order to have it at no charge.

Remember this:

Oxidizing agent (oxidant) = species that is reduced
Reducing Agent (Reductant) - species that is oxidized

 

[thejamespark]

I am unsure of how to answer this question and I don't understand the answer

1. When the volume of a gas is decreased at constant temperature, the pressure increases because the molecules:
a) Move faster
b) Move slower
c) Become heavier
d) Become lighter
e) Strike a unit area of the container more often

The answer is e)
Doesn't decreasing the volume = decrease in the gas molecules. Wouldn't that lead to more dispersion of the gas molecules?

2. The following is a spontaneous oxidation-reduction reaction:
(Cr2O7^2-) + (14H^+) + (6I^-) + --> (2Cr^3+) + 7H2O + 3I2
Which of the following is the best reducing agent?
The answer is I- but I don't understand why H+ the best reducing agent


Thanks in advance

I will try to make it short and concise.

1)If you decrease the volume, pressure will increase.
As the person above me stated it, if you have 10 people in a large room and you move those 10 people to a small bathroom what happens? The pressure increases.
If you increase the pressure, the molecules that are constantly moving around would hit the walls of the container more frequently because its AVERAGE velocity of those molecules increased due to increased pressure.

2) (Cr2O7^2-) + (14H^+) + (6I^-) + --> (2Cr^3+) + 7H2O + 3I2

For this keep it simple:
Reducing agent(reductant) = oxidizing, LEO
Oxidizing agent(oxidant) = reducing, GER

Look at the only two Cr207^-2 and 6I-
From Cr207^-2 :
[O7] = (-2) x 7 = -14
Since the O7 = -14, the total reaction must add up to 0, which makes the Cr2 = to +12 (14-2) since the total reaction is ^-2
[Cr2] = Therefore, it is +12 (+6 x 2Cr)

If you look at the reaction from reactants -> product side, CR has gone from +6 to +3. This Cr207 has undergone Reduction(oxidizing agent).
And if you look at the I-, it has gone from -1 -> 0(since I2 = standard elemental state = 0) which is: Oxidizing(Reducing agent)
Therefore, we will say that I- is the Reducing Agent(Oxidizing).

I hope this helps....

 

[Pdentalstudent]

thanks guys!!! I get it now!! that people analogy is great!

Thanks